Answer:
There is 37.36 grams of K3PO4 produced
Explanation:
Step 1: Data given
Mass of H3PO4 = 29.6 grams
KOH is in excess
Molar mass of KOH = 56.11 g/mol
Molar mass of H3PO4 = 97.99 g/mol
Step 2: The balanced equation
3KOH(aq) + H3PO4(aq) ⇔ K3PO4(aq)+3H2O(l)
Step 3: Calculate mass of KOH
Mass KOH = mass KOH / molar mass KOH
Mass KOH = 29.6 grams / 56.11 g/mol
Mass KOH = 0.528 moles
Step 4: Calculate moles of K3PO4
Since KOH is the limiting reactant, We need 3 moles of KOH for each moles of H3PO4, to produce 1 mole of K3PO4 and 3 moles of H2O
For 0.528 moles of KOH we'll have 0.528/3 = 0.176 moles of K3PO4
Step 5: Calculate mass of K3PO4
Mass K3PO4 = moles K3PO4 * molar mass K3PO4
Mass K3PO4 = 0.176 moles * 212.27 g/mol
Mass K3PO4 = 37.36 grams
There is 37.36 grams of K3PO4 produced
